Tipler_Llewellyn

11-2 Ground-State Properties of Nuclei 491 (a) 0.5 × 10–6 eV (b) Figure 11-11 (a) Transitions 2P3/2 2 × 10–3 eV F=3 between the sodium doublet 2P1/2 F=2 F=1 levels produce the yellow D 2.1 eV F=0 lines, 2P1>2 S 2S1>2 being D1 and 2P3>2 S 2S1>2 being D2 . 2S1/2 F=2 Coupling between the F=1 atomic angular momentum J 0.8 × 10–6 eV and the nuclear spin I ϭ 3>2 results in the hyperfine F=2 7.3 × 10–6 eV splitting, each level having total angular momentum F ϭ I ϩ J. Note that the hyperfine splitting of each of the doublet levels is about 10Ϫ3 times that of the fine-structure splitting of the 2P level. (b) The selection rule ¢F ϭ Ϯ1, 0 leads to the D2 line being split into six components. The D1 line is correspondingly split into four components (not shown). F=1 Evidently the nucleons couple together in such a way that their angular momenta add The degeneracy of the to zero in pairs, as is often the case for electrons in atoms. There is no such simple rule hyperfine levels in nuclei with for other nuclides with either odd N or odd Z or both. Some of the successes of the nonzero spins, e.g., the shell model to be discussed in Section 11-6 are the correct prediction of nuclear spins proton in 1H, is removed by for many nuclei. an external B field, a nuclear analog of the Zeeman effect. The magnetic moment of the nucleus gNmI␮N is of the order of the nuclear Transitions between these magneton, ␮N ϭ eU>2mp , since the magnitude of gN is typically between 1 and 5 levels, separated (in 1H) by 2␮pB, oscillate at the spin and the maximum value of ƒm1 ƒ ϭ I. The exact value is difficult to predict because precession rate. Detection of the resulting absorption or it depends on the detailed motion of the nucleons. If the proton and neutron obeyed emission of radiation allows the Dirac relativistic wave equation, as does the electron, the magnetic moment due “mapping” of the hydrogen- to spin would be 1 nuclear magneton for the proton because its charge is ϩe and 0 containing soft tissue, the for the neutron because it has no charge.11 The experimentally determined moments basis for medical magnetic of the nucleons are resonance imaging (MRI). (␮p)z ϭ ϩ2.79285 ␮N (␮n)z ϭ Ϫ1.91304 ␮N As we will see in Chapter 12, the proton and neutron are more complex particles than the electron. It is interesting that the deviations of these moments from those predicted by the Dirac equation are about the same magnitude, 1.91 for the neutron and 1.79 for the proton. The reason that the magnetic moments of the nucleons have these partic- ular values is not yet completely understood; the current theoretical predictions of ␮p and ␮n agree with high-precision, experimentally measured values only to within about 1 percent.

492 Chapter 11 Nuclear Physics EXAMPLE 11-7 Nuclear Spin of Thallium-205 High-resolution spectroscopic study of the spectrum of 205Tl reveals that each component of the doublet 2P1>2 S 2S1>2 (377.7 nm), 2P3>2 S 2S1>2 (535.2 nm) consists of three hyperfine components. This requires that there be two hyperfine levels for each J. Determine the spin of the 205Tl nucleus. SOLUTION If I Յ J, then there are (2I ϩ 1) different F levels, and if I Ͼ J, there are (2J ϩ 1) different F levels. Since the hyperfine spectrum indicates that there are two levels for each J, then for the 2P3>2 level either 2I ϩ 1 ϭ 2 or 2J ϩ 1 ϭ 2 But we already know that J ϭ (3>2), so (2J ϩ 1) cannot equal 2; therefore (2I ϩ 1) ϭ 2 and the spin of the 205Tl nucleus (in its ground state) must be 1>2. Note that for the 2P1>2 and 2S1>2 levels both of the equations above are satisfied since in these two cases I ϭ J. Questions 1. Why is N approximately equal to Z for stable nuclei? Why is N greater than Z for heavy nuclei? 2. Why are there no stable isotopes with Z Ͼ 83? 3. The mass of 12C, which contains 6 protons and 6 neutrons, is exactly 12.000 u by the definition of the unified mass unit. Why isn’t the mass of 16O, which contains 8 protons and 8 neutrons, exactly 16.000 u? 11-3 Radioactivity Of the more than 3000 nuclides known, only 266 are stable. All of the rest are ra- dioactive; that is, they decay into other nuclides by emitting radiation. The term radi- ation here refers to particles as well as electromagnetic radiation. In 1900 Rutherford discovered that the rate of emission of radiation from a substance was not constant but decreased exponentially with time. This exponential time dependence is characteris- tic of all radioactivity and indicates that it is a statistical process. Because each nu- cleus is well shielded from others by the atomic electrons, pressure and temperature changes have no effect on nuclear properties.12 For a statistical decay (in which the decay of any individual nucleus is a random event), the number of nuclei decaying in a time interval dt is proportional to dt and to the number of nuclei present. If N(t) is the number of radioactive nuclei at time t and ϪdN is the number that decay in dt (the minus sign is necessary because N de- creases), we have ϪdN ϭ ␭N dt 11-17 where the constant of proportionality, ␭, is called the decay constant. ␭ is the proba- bility per unit time of the decay of any given nucleus. The solution of this equation is N(t) ϭ N0eϪ␭t 11-18